Dynamics Problem about Velocity

[joryboy]

Homework Statement

Hi guys, I was working on this problem and I want to know if it can be solved by using an indefinite integral, so here it goes:

A car starts from rest and travels along a straight track such that it accelerates at 10 m/s^2 for 10 s, and then decelerates at 2 m/s^2 . Determine the velocity of the car.

Homework Equations

My question is if I can solve this using INDEFINITE INTEGRALS ONLY?

The Attempt at a Solution

I have obtained the correct answer using a definite integral:

For 0 less than or equal to t less than or equal to 10:
Integral from 0 to v (1) (dv) = integral from 0 to t (10) dt v=10t
For 10 less than or equal to t less than or equal to t'
Integral from 100 to v (1) dv = integral from 10 to t (-2 ) dt = v=-2t+120

I'm able to obtain the same result (when using an indefinite integral) for t between 0-10 but for 10-t I cant, also in my book it says that v1 = delta v + v0, well I can clearly see that when using the definite integral but not when I have to use the indefinite integral..

I tried this for t between 10-t : int (1) dv = int (-2) dt = v+c = -2t+C = v=-2t which is incorrect, then I said ok I forgot to add v0 but still I get v=-2t+100 which is still incorrect..

Also if I see it like this: v1= -2t + C + v0 I still can't get a clear understanding because I can't get a value for C.. Please help me and thanks!

 

[Simon Bridge]

My question is if I can solve this using INDEFINITE INTEGRALS ONLY?

This question should be done by indefinite integration.

The velocity is provided by the area under the acceleration-time graph.
For the first section: $$v(0<t\leq 10)=10\int \;dt = 10t+c$$ ... here, $c$ is the constant of integration
- the indefinite integral gives you a family of possible solutions, only one of them is the solution to the problem you have. Find it by working out which value of c matches your problem.

- to work out what $c$ has to be, you use the fact that you also know the value of v at a particular time t ... specifically at t=0. So you put that point into the general solution and solve for c.

For the second section $$v(t>10) = -2 \int\; dt = -2t+d$$ ... here $d$ is the constant of integration and $v(0) \neq 0$ because this part of the journey did not go through $t=0$.
What you need to find $d$ is some point on the line for the second part of the journey you already know the value of.